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\section[Operator relations and one-dimensional dynamical\break
 systems]{Operator relations and one-dimensional dynamical systems}
\label{sec:2.1}

\subsection{Operator relations connected with one-dimen\-sional  dynamical
systems}\label{sec:2.1.1}
\markright{2.1. One-dimensional dynamical systems}

\noindent\textbf{1.}
Consider an operator $X$ satisfying, together with its adjoint
$X^*$, an algebraic relation of the form
\begin{equation}
\label{xx}
XX^* = F(X^*X),
\end{equation}
where $F(\cdot)\colon \mathbb{R} \to \mathbb{R}$ is a mapping measurable
with respect to the Borel $\sigma$\nobreakdash-algebra.

If $F(\cdot)= P_n(\cdot)$ is a real polynomial, then the pair of
operators $X$, $X^*$ satisfying relation \eqref{xx} is a
representation of a $*$-algebra $\mathcal{A}$ generated by elements
$x$, $x^*$ satisfying the relation $xx^* = P_n(x^*x)$. In some
non-polynomial cases,
pairs of operators $X$, $X^*$ form representations of $C^*$-algebras
or other topological algebras, but we will not concentrate our
attention on the underlying algebraic objects, restricting ourselves
to the study of representations of the relation.

We have already considered examples of representations of the form
\eqref{xx}
above. For the Hermitian $q$-plane, $xx^* = q x^*x$, $q \in \mathbb{R}$,
we have $F(\lambda) = q\lambda$, and for $q$-CCR, $xx^* = qx^*x +1$, $q
\in \mathbb{R}$, we have $F(\lambda) = q\lambda +1$.

Below we consider other examples of relations of the form \eqref{xx}.

\begin{example} (Second degree mappings). Consider the following relation
\begin{equation}\label{rel:quadratic}
xx^* = \alpha x^*x(I- x^*x), \qquad a >0.
\end{equation}
The corresponding polynomial is $F(\lambda) = \alpha
\lambda(1-\lambda)$. Any
second-degree mapping can be reduced to such a form by using an affine change of
variables; however, positivity considerations that will be discussed below make the
representation problem different for different forms of the polynomial.
Here, we consider only relations \eqref{rel:quadratic} and
\begin{equation}\label{rel:parabola}
xx^* = (x^*x - q I)^2
\end{equation}
that corresponds to the polynomial $\tilde F(t) = (t - q)^2$.
The two latter polynomials can be transformed to each other by the
change of variables $t = - \alpha \lambda + b$, with $\alpha
= 1+ \sqrt{4q+1}$.
The structure of bounded representations of such relations greatly depends
on the value of $\alpha$ (see below).
\end{example}

\begin{example} (Continued fractions).
Another interesting class of relations related to continuous fractions
is given by the following relations
\begin{gather*}
xx^* = (a x^*x +c)(bx^*x +d)^{-1},
\\
a,b,c,d \in \mathbb{R},\quad
b\ne 0, \quad ad -bc \ne 0.
\end{gather*}
Below, we will show how representations of the relation depend on
the values
of the parameters.

In particular, the two-parameter unit quantum disk algebra \cite{klles}
is generated by the relation
\begin{gather*}
qzz^* - z^*z = q-1 + \mu (1-zz^*)(1-z^*z),
\\
 0\le \mu \le 1, \quad 0\le
q \le 1, \quad (\mu, q) \ne (0,1),
\end{gather*}
which can be rewritten in the form \eqref{xx} with
\[
F(\lambda) = \frac{(q+\mu) \lambda + 1 -q - \mu}{\mu \lambda + 1 -
\mu}.
\]
\end{example}

\begin{remark}
As we will see, in many examples unbounded operators $X$ naturally arise as
representations of such a relation. Therefore, it is necessary
to accurately define the meaning of the relation for unbounded operators.
Occasionally, derived formulas will make
sense for unbounded operators, too, or will yield unbounded
operators; however,
we will not
discuss problems related to unbounded
operators here.
\end{remark}

\medskip \noindent \textbf{2.}
Let $X$ be bounded. For a polar decomposition of the operator $X$,
we have $X= UC$, where $C = C^* = (X^*X)^{1/2}$, $U$ is a partial
isometry, and $\ker U= \ker C = \ker X$, $U^*U$ is an orthogonal
projection onto $(\ker C)^\perp$. Taking into
account relation \eqref{xx}, we get
\begin{equation} \label{ucu}
UC^2U^* = F(C^2),
\end{equation}
which implies that
\begin{equation} \label{cu}
UC^2 = F(C^2) U, \quad C^2 U^* =U^* F(C^2).
\end{equation}
Notice that \eqref{ucu} implies that
$ \ker U^* = \ker X^* =\ker F(C^2)$ as well.

\medskip \noindent\textbf{3.}
 Now we establish some relations which follow from \eqref{xx}. First
notice that $X(X^*X) = F (X^*X) X$ and $(X^*X) X^* = X^* F(X^*X)$.
For any $k=1$, 2,\dots, we recursively get the relations $X(X^*X)^k
=(F(X^*X))^kX$, and $(X^*X)^k X^* = X^* (F(X^*X))^k$. From the two latter
relations we get $XP(X^*X) = P(X^*X)X$
for polynomials of $X^*X$, and in the case of bounded
operators, a standard approximation procedure using the spectral
decomposition of the self-adjoint operator $X^*X$ yields
\begin{equation}\label{xx_fun}
X \phi(X^*X) = \phi(F(X^*X)) X, \quad \phi(X^*X) X^* = X^*
\phi(F(X^*X))
\end{equation}
for any bounded measurable function $\phi(\cdot)$. In particular,
\eqref{xx_fun} implies that for all $n=1$, 2,~\dots,
\begin{align} \label{xx_n}
X^n \phi(X^*X) &= \phi(F^{ n}(X^*X) X^n,
\notag
\\ \phi(X^*X) (X^*)^n
&=(X^*)^n\phi(F^{ n}(X^*X)),
\end{align}
where $F^{ n}(\cdot)$ is the $n$-th iteration of $F(\cdot)$.

We have the following statement.

\begin{proposition}\label{cu-props}
Let $X$ be a bounded operator satisfying \eqref{xx} with a bounded
measurable real function $F(\cdot)$, and let $X=UC$ be the polar
decomposition introduced above,  such
that $\ker C^2 = \ker U$, $\ker F(C^2) =
\ker U^*$.
The relation \eqref{xx} is
equivalent to any of the following\textup:

\begin{itemize}

\item[a\textup)] $C^2 U^* = U^* F(C^2)$\textup;

\item[b\textup)] $X \phi(X^*X) = \phi(F(X^*X)) X$,
for any  measurable function $\phi(\cdot)$ bounded on the spectrum
of $X^*X$\textup;

\item[c\textup)]
$\phi(C^2) U^* = U^* \phi(F(C^2))$,  for any measurable function $\phi(\cdot)$ bounded on the
spectrum of $X^*X$\textup;

\item[d\textup)]  $E_{C^2}(\Delta)
U^* = U^* E_{C^2}(F^{-1}(\Delta))$ for all
measurable $\Delta\subset \mathbb{R}$\textup.

\end{itemize}
\noindent
Here $E_{C^2}(\cdot) $ is a resolution of the identity of the
operator $C^2$, and $F^{-1}(\Delta)$ denotes the full pre-image of
$\Delta$ \textup(more precisely, the full pre-image of $\Delta \cap
F(\mathbb{R})$\textup).
\end{proposition}

\begin{proof}
It was already shown that \eqref{xx} is equivalent to a), and that
\eqref{xx} implies b). To show that b) implies \eqref{xx}, take
$\phi(\lambda) =\lambda$. Multiplying by $X^*$ from the right, we
have $XX^*XX^* = F(X^*X) XX^*$, which implies that  relation
\eqref{xx} holds
on vectors orthogonal to $\ker XX^*$. On the other
hand, $\ker X^* = \ker U^* = \ker F(X^*X)$.

The proof that c) is equivalent to a) can be done in the same way as b)
was derived. Relation d) is a particular case of c) with $\phi(\lambda) =
\chi_\Delta(\lambda)$ (notice that $E_{F(C^2)}(\Delta) =
E_{C^2}(F^{-1}(\Delta))$);
 using a spectral
representation for measurable functions of the positive self-adjoint
operator $C^2$, $\phi(C^2) = \int \phi(\lambda)\, dE_{C^2}(\lambda)$,
one can easily see that c) follows from d)
as well.
\end{proof}

\noindent\textbf{4.}
Operators of the form \eqref{xx} form a subclass in the class of
 centered
operators. Recall that a bounded operator $X$ is centered, if the
operators $X^l(X^*)^l$, $(X^*)^kX^k$, $k,l=1$, 2,\dots, form a
commuting family.

\begin{proposition}
 \label{xx-center}
A bounded operator $X$ satisfying \eqref{xx} is centered. If a pair
of operators, a self-adjoint $C$ and a partial isometry $U$, such that
$\ker U =  \ker C^2$, $\ker U^* = \ker F(C^*)$, satisfy
relation \eqref{cu}, then $U$ is a centered partial isometry.
\end{proposition}


\begin{proof}
Let us show that $X$ is centered. Write $A_k =X^k(X^*)^k$, $B_l = (X^*)^lX^l$,
$k,l\ge 1$. First consider the operators $A_k$. Applying \eqref{xx_n} we
have
\begin{align*}
A_k &= X^k(X^*)^k= X^{k-1} (XX^*) (X^*)^{k-1}
\\
&= X^{k-1}F(X^*X) (X^*)^{k-1} = F^{ n} (X^*X) X^{n-1}(X^*)^{n-1}
\\
& = F^{ n} (X^*X) X^{n-2}F(X^*X)(X^*)^{n-2}
\\
& = F^{ n} (X^*X) F^{ n-1} (X^*X)\dots F(X^*X).
\end{align*}
Since all the operators $A_k$, $k\ge 1$, are functions of the  single
operator $X^*X$, they commute with each other.

Now consider a pair $A_k$, $B_l$. Again, applying \eqref{xx_n} and
the obtained representation for $A_k$, we get
\begin{align*}
B_l A_k &= (X^*)^l X^l X^k (X^*)^k
\\
&= (X^*)^l X^l F^{ k} (X^*X)
F^{ k-1} (X^*X)\dots F(X^*X)
\\
&= (X^*)^l F^{ k+l} (X^*X)
F^{ k+l-1} (X^*X)\dots F^{l+1}(X^*X) X^l
\\
&= F^{ k} (X^*X)
F^{ k-1} (X^*X)\dots F(X^*X) (X^*)^l X^l= A_k B_l,
\end{align*}
and therefore, the operators $A_k$ and $B_l$, $k, l \ge 1$, also
commute with each other.

Now we show that the operators $B_l$, $l\ge 1$, also commute with each
other. Indeed, for $k>l$, we have
\begin{align*}
B_k B_l &= (X^*)^k X^k (X^*)^l X^l =(X^*)^l (X^*)^{k-l} X^{k-l}  X^l
(X^*)^l X^l
\\
&= (X^*)^l B_{k-l} A_l X^l = (X^*)^l A_l B_{k-l} X^l
\\
&=
(X^*)^l X^l (X^*)^l (X^*)^{k-l} X^{k-l}  X^l = B_l B_k.
\end{align*}
Thus, the operator $X$ is centered.

It remains to prove the second statement of the proposition. Let
$\Delta  = \sigma(C^2) \setminus \{0\}$. Then $E_{C^2}(\Delta)$ is
a projection onto the co-kernel of $C^2$, and, since $\ker U = \ker C$,
we have $E_{C^2}(\Delta) = U^*U$. Then \eqref{cu} implies that $UU^* =
UE(\Delta)U^* =  E_{C^2}(F^{-1}(\Delta))$. Similarly,
\begin{align*}
U^k (U^*)^k &= U^{k-1} U U^* (U^*)^{k-1}= U^{k-1} U U^* U U^*
(U^*)^{k-1}
\\
&= U^k E_{C^2}(\Delta) (U^*)^k = E_{C^2} (F^{-k}(\Delta)).
\end{align*}
Therefore, the projections
$U^k (U^*)^k$, $k\ge1$, commute both with $U^*U$
and each other. The rest of the commuting relations can be obtained in
the same way as it was done for $X$ in the proof of the first part of
the proposition.
\end{proof}

\noindent\textbf{5.}
Now we will show that properties of representations of the relation
 \eqref{xx} depend on properties of the corresponding dynamical
system $\lambda \mapsto F(\lambda)$, and study this dependency in
details.

\begin{proposition}\label{toeigen}
Let the operators $C$ and $U$ satisfy relation \eqref{cu}. If
$e_\lambda$ is an eigenvector  of the operator $C^2$, corresponding to an eigenvalue
$\lambda$, then $U^* e_\lambda$ is either zero or is again
an eigenvector of
$C^2$,
corresponding to the eigenvalue $F(\lambda)$.
\end{proposition}

\begin{proof}
Indeed, from  \eqref{cu}, we see that $C^2
U^* e_\lambda = U^* F(C^2)\,
e_\lambda = F(\lambda)\, U^* e_\lambda$.
\end{proof}

In particular, if the operator $U$ is unitary, then $U^*e_\lambda$ is
not zero, which implies that, in this case, $F(\cdot)$ maps the spectrum
of $C^2$ into itself.

A similar fact holds for all points of the spectrum of the operator
$C^2$.

\begin{proposition}
If the operator $X$ in \eqref{xx} is invertible \textup(i.e., $U$ in
\eqref{cu} is unitary\textup),
then $F(\cdot)$ maps the spectrum $\sigma(C^2)$
into itself.
\end{proposition}

\begin{proof}
Let $\lambda \in \sigma(C^2)$. Then there exists a sequence of unit
vectors $e_n \in H$, $n\ge 1$, such that $\|C^2 e_n - \lambda e_n\| \to
\infty$, $n \to \infty$. Then
\begin{gather*}
\|C^2 U^* e_n - F(\lambda) U^* e_n\| = \|U^*(F(C^2) e_n - F(\lambda)
e_n)\|
\\
= \|F(C^2) e_n - F(\lambda) e_n\| \to 0, \qquad n\to \infty,
\end{gather*}
and $\|U^* e_n\|=1$, $n \ge 1$. This implies that $F(\lambda )\in
\sigma(C^2)$.
\end{proof}

Proposition \ref{toeigen} provides a way for constructing representations
of relation \eqref{xx}. Indeed, choose a sequence of positive numbers
$\lambda_k$,  $k \in \mathbb{Z}$, (if
it exists) such that $F(\lambda_k) = \lambda_{k+1}$ for all $k \in
\mathbb{Z}$, and define the operators $C$, $U$ in $l_2(\mathbb{Z})$ as
follows:
\begin{equation} \label{rel:chains}
C^2 e_k = \lambda_k, \quad U e_k = e_{k-1}, \qquad k \in \mathbb{Z}.
\end{equation}
Then the operators $U$, $C$, satisfy \eqref{cu}, and therefore, give a
representation of \eqref{xx}. Below, we will study whether other
irreducible representations exist and what is their structure. In
particular, for ``simple'' mappings $F(\cdot)$, all irreducible
representations with unitary $U$ have such a form, and can be classified up
to a unitary equivalence.

\medskip\noindent \textbf{6.}
Let
a Borel  set $\Delta$ be such that $\Delta \subset F(\mathbb{R})$ and
$F^{-1}(\Delta) = \Delta$.
Proposition~\ref{cu-props} implies that $E_{C^2}(\Delta) U^* = U^*
E_{C^2}(F^{-1}(\Delta)) = U^* E_{C^2}(\Delta)$, i.e., $E_{C^2}(\Delta)$
is a projection onto an invariant subspace in $H$. Therefore, to study
irreducible representations, we need to study the ``smallest''
invariant
subsets (in the sense indicated above) that contains the spectrum of $C^2$. 
Below we will use some facts about dynamical systems and their invariant 
sets, as well as properties of the corresponding spectral measures.

\medskip\noindent\textbf{7.}
Recall some basic notions and facts about discrete dynamical systems.
A discrete time one-dimensional dynamical system
\index{dynamical system}
is just a
(continuous or measurable) mapping ${\mathbb{R}^{1}}\ni\lambda
\mapsto F(\lambda)\in{\mathbb{R}^{1}}$.

First of all, we recall the notion of a trajectory or
an orbit of a dynamical
system.
Traditionally, a trajectory or an orbit
\index{orbit}
\index{dynamical system, orbit of}
of the dynamical system
$F(\cdot):{\mathbb{R}^{1}}\mapsto{\mathbb{R}^{1}}$
is the set 
$$\orb(\lambda)=\{\lambda,F(\lambda),F^{2}(\lambda),\dots\}=
\bigcup_{n=0}^{\infty} {F^{n}(\lambda)}.
$$
 Here
$F^{n}(\cdot)=F(F^{n-1}(\cdot))$, $n=1$, 2,~\dots, and $F^{0}(\cdot)$
is the identity transformation. Since we are interested in the action
of $F^{-1}$ as well, will say that two points, $\lambda_1$ and
$\lambda_2$,
belong to the same trajectory, if $F^k(\lambda_1) =
\lambda_2$ or $F^k(\lambda_2)
= \lambda_1$ for some $k \ge 0$. We also need the notion of the trajectory
decomposition: we say
that two
points, $\lambda_1$ and $\lambda_2$, belong to the same
element of the trajectory decomposition, if these points ``meet
in the future'', i.e., if $F^k(\lambda_1)
= F^m(\lambda_2)$
for some $k$, $m>0$.
A point $\lambda_{0}\in{\mathbb{R}^{1}}$,
such that $F^{m}(\lambda_{0})=\lambda_{0}$ and $F^{n}(\lambda_{0})\neq
\lambda_{0}$
for $0<n<m$, is called a periodic point
\index{periodic point}
\index{dynamical system, periodic point of}
of period $m$.
The periodic points $\lambda_{0}$,
$F(\lambda_{0})$, \dots, $F^{m-1}(\lambda_{0})$
form a cycle
\index{cycle}
\index{dynamical system, cycle of}
of period~ $m$.

Bearing in mind the  representations of relation \eqref{cu} and the fact
that these invariant sets should carry the spectral measure of the
operator $C^2$, we need to
consider a measurable
mapping of a measurable space. In this case, the mapping
$F(\cdot)$ gives rise to a mapping of Borel measures on the
line by the formula
\[
d\rho(\lambda) \mapsto d\rho(F^{-1}(\lambda)).
\]

A Borel measure $\rho(\cdot)$ is called quasi-invariant with
respect to $F(\cdot)$, if $\rho(F^{-1}(\cdot))$ is absolutely
continuous with respect to $\rho(\cdot)$.

\begin{proposition}
If the operator $U$ in \eqref{cu} is unitary, and $F(\cdot)$ is
one-to-one, then the spectral measure of\/ $C^2$ is quasi-invariant with
respect to $F(\cdot)$ and $F^{-1}(\cdot)$.
\end{proposition}

\begin{proof}
Indeed, we can take the spectral measure to be in the form $\rho(\cdot) =
(E_{C^2}(\cdot)\omega, \omega)$, where $\omega$ is a vector of maximal spectral
type. Then the statement follows from the equality
\begin{align*}
\rho(F^{-1}(\cdot)) &= (E_{C^2}(F^{-1}(\cdot))\omega, \omega)
\\
& = (U^*
E_{C^2}(F^{-1}(\cdot))
\omega, U^*\omega) = (E_{C^2}(\cdot) U^* \omega, U^*\omega)
\end{align*}
and a similar relation for $U$.
\end{proof}

We also need the notion of ergodicity of a measure. A Borel
measure $\rho(\cdot)$ is called ergodic with respect to the action of
a dynamical system $F(\cdot)$, if for any measurable
$F(\cdot)$-invariant set $\Delta\subset\mathbb{R}$, (i.e., the
set $\Delta$ such that $F^{-1}(\Delta) = \Delta$), we have that either
$\rho(\Delta)=0$ or $\rho(\mathbb{R}\setminus \Delta)=0$.

\begin{proposition}
If a pair $X$, $X^*$, satisfying \eqref{xx}, is irreducible, then the
spectral measure of the operator\/ $C^2$ is ergodic with respect to
$F(\cdot)$.
\end{proposition}

\begin{proof}
Indeed, otherwise there would exist a non-trivial invariant subset
$\Delta\subset\sigma(C^2)$. As was noticed above, in this case
$E_{C^2}(\Delta)$ is a projection onto an invariant subspace which is
non-trivial
if and only if the spectral measure of $\Delta$ is neither zero
nor one.
\end{proof}

The simplest invariant sets are elements of the trajectory
decomposition
of the dynamical
system (in the one-to-one case, they are just orbits).
The simplest class of quasi-invariant
ergodic measures is formed by atomic measures concentrated on an
element of trajectory decomposition; however, an atomic measure
concentrated on an orbit is also quasi-invariant and ergodic. Below, we
will see that only such measures corresponding to an orbit give rise to
irreducible representations of the relation.

The existence of non-atomic
quasi-invariant ergodic measures depends on topological
properties of the dynamical system. In the one-to-one case, we
have the following
fact.

\begin{proposition}
If a dynamical system $\lambda\mapsto F(\lambda)$
with one-to-one $F(\cdot)$
possesses a measurable
section, i.e., a measurable set that intersects any orbit
in a single point, then any ergodic measure is concentrated
on a single orbit of the dynamical system.
\end{proposition}

In the non-bijective case, the condition of existence
of a measurable section is replaced by the
following condition of existence of $M$-partition:

{\em
the dynamical system possesses an $M$-partition, i.e., there exists a
partition $\mathbb{R} = \bigcup_{n \in \mathbb{N}} \Delta_n$, $\Delta_k
\in \mathfrak{B}(\mathbb{R})$, such that

\textup{1)} for any $k$ there exists $j$ such that
$F(\Delta_k)=\Delta_j$ and $F(\cdot)$ is one-to-one on
$\Delta_k$\textup{; }

\textup{2)} if for some $n$\textup{,} $k$, the mapping $F^n(\cdot)$ maps
$\Delta_k$ into itself\textup{,} then $F^n(\cdot)$ is the identity on
$\Delta_k$.}

We also mention the following statement from \cite{vai_fed}.

\begin{theorem}\label{th:per}
Let $I$ be some finite interval, and $F(\cdot)\colon I \to I$ be a
continuous partially monotone mapping \textup(i.e.,  $I$ decomposes into a
finite union of sub-intervals, on which $F(\cdot)$ is monotone\textup).
Then
the following conditions are equivalent\textup:

i\textup)
there exists an $M$-partition of $I$, each element of which is an
interval \textup(possibly, a single point\textup)\textup;

ii\textup) any quasi-invariant ergodic measure is concentrated on a single
element of the trajectory decomposition\textup;

iii\textup) the set of periodic points, $\per F$, is closed\textup;

iv\textup)
for some $m\ge 0$, the relation $\fix(F^{2^{m+1}}) = \fix(F^{2^m})$
holds \textup($\fix (F)$ denotes the set of fixed points of $F$\textup).
\end{theorem}

Below we study the correspondence between the orbits and
irreducible  representations of the relation.

\subsection{Finite-dimensional representations}\label{sec:2.1.2}

If
the sequence $\lambda_k$ in \eqref{rel:chains} is periodic, the
corresponding irreducible representation is finite-dimensional.
We will show here that all finite-dimensional representations of relation
\eqref{xx} are related to cycles of the corresponding dynamical system.
Then we apply the Sharkovsky theorem on existence of cycles to
to study irreducible finite-dimen\-sional representations; these
results are illustrated with examples.

\medskip\noindent\textbf{1.}
Let us classify
irreducible pairs $X$, $X^*$ of operators on a finite-dimensional
space, obeying  relation \eqref{xx} 
 (irreducible
finite-dimen\-sional representations of \eqref{xx})  up
to unitary equivalence.

\begin{theorem} \label{fin_th}
Any cycle $O_\lambda=\{\lambda, F(\lambda), \dots , F^{n-1}(\lambda)\}$ of
period
$n$, such that $\lambda \ge0$ and $t>0$ for all other points $t
\in O_\lambda$,
defines a family of $n$-dimensional irreducible representations
of \eqref{xx}\textup:
\begin{equation}\label{one5}
X=
\begin{pmatrix}
0&\sqrt{F(\lambda)}&&0\\
\vdots&\ddots&\ddots&\\
0 &&0&\sqrt{F^{n-1}(\lambda)}\\
e^{i\phi}\lambda&0&\dots&0\end{pmatrix},
\end{equation}
where $\phi\in [0,2\pi)$, if $\lambda>0$, or a single representation,
if $\lambda=0$.

These are all, up to unitary equivalence, distinct irreducible
representations of  \eqref{xx}.
\end{theorem}

\begin{proof}
Indeed, by a direct calculation one can check that the representation
\eqref{one5} satisfies the necessary relation. Let us show that it
is irreducible. Any bounded self-adjoint operator $T$,  which
commutes with $X$ and $X^*$, commutes with 
$C^2$ and $U$ 
introduced  above as well. But in the selected basis, $C^2$  is
diagonal with distinct eigenvalues $\lambda$, \dots,  $F^{n-1}(\lambda)$.
Then $T$ is diagonal, and the commutation with $U$ implies that $T= cI$.


We show that these are all the irreducible representations of
\eqref{xx}.

Assume that $U$ is unitary.
Since $\dim H < \infty$,
the operator $C^2$ can be diagonalized,
and has non-negative eigenvalues.
Take an eigenvalue $t$ and a  unit eigenvector $e_t$ of $C^2$,  $C^2
e_t = t e_t$. It follows from
\eqref{cu} that $U^*e_t=e_{F(t)}$ is  again a unit eigenvector of $C^2$,
$C^2U^*e_t = F(t) e_{F(t)}$. Consider the sequence of
unit vectors $(U^*)^ke_t$, $k\ge 0$.  Since
there is only a finite number of eigenvalues  of
$C^2$, we conclude that, for some $n\ge1$ and  some eigenvalue
$t_0$,
$F^n(t_0) = t_0$. The linear span $H_0$ of eigenspaces  of
$C^2$, corresponding to eigenvalues $t_0$, \dots,  $F^{n-1}(t_0)$,
is invariant with respect to $C^2$ and $U^*$.

We show that the space $H_0$ is  also invariant with respect to $U$.
Assume that $Ue_{t_0}$ does not belong to $H_0$.  Denote
by $u$ the  projection
of $Ue_{t_0}$ on the orthogonal complement of $H_0$.
From the relation $E_{C^2}(\Delta) U^* = U^*E_{C^2}(F^{-1}(\Delta))$ for
any measurable $\Delta$, we
have that $u $ belongs to the space $H_1$ generated by eigenvectors
corresponding to all eigenvalues $\lambda\in \Lambda = \{ F(\lambda) =
t_0, \lambda \ne F^{n-1}(t_0)\}$.

Since any point has a single image, for
all $k>0$, the sets $\Lambda_k = F^{-k}(\Lambda)$
are disjoint. We
have that $U^k H_1 \subset E_{C^2}(F^{-k}(\Lambda)) H_1$,  and
all these spaces are nonzero and orthogonal to each  other.
But this is impossible, since $H$ is finite-dimensional.
Therefore, $H_0$ is invariant with respect to $X$ and  $X^*$,
and, due to the irreducibility, coincides with the whole of $H$.

The operator $U^n$ commutes with $C^2$, since  $F^n(t_0)
=t_0$. It also obviously commutes with $U$ and
$U^*$. Therefore, the irreducibility implies  $U^n
= \alpha I$, $|\alpha|=1$. In the basis $e_t$,  \dots,
$e_{F^{n-1}(t)}$, the operators act as needed.


Consider the case of non-unitary $U$. Now, $\ker C^2 =\ker U \ne
\{0\}$, and there
exists a unit vector $e_0$ such that $Ue_0=C^2 e_0=0$. Again, consider
the vectors $(U^*)^k e_0$, $k=0$, 1,~\dots. Relation \eqref{cu} implies that
$e_k$ is either an eigenvector of $C^2$ with the eigenvalue $F^k(0)$, or the zero
vector. Consider two possibilities: there exists $n$ such that $e_k=0$,
or else there are numbers $k$ and $n$, $k<n$, such that $F^n(0) =
F^k(0)$.

We show that the second alternative is impossible. Indeed, since $\ker U^*
= \ker F(C^2)$, we conclude that the multiplicity of the eigenvalue $F^k(0)$ is
the same as that of $F^n(0)$. On the other hand, two orthogonal
eigenspaces,
$H_{k-1}$ and $H_n$ that correspond
to $F^{k-1}(0)$ and $F^n(0)$, are mapped by $U^*$ into the eigenspace
$H_k$ corresponding to $F^k(0)$. Since the space is finite-dimensional, there exists a
non-zero vector $w$ in
the direct sum $H_{k-1} \oplus H_n$ such that
$U^* w=0$, which contradicts the condition $\ker U^* = \ker F(C^2)$.

Then, there exists $n$ such that $e_k=0$,
$k\ge n$, i.e., $e_{n-1} \in \ker U^*=\ker F(C^2)$, and $F^n(0)=0$; thus
0 is a periodic point of period $n$, and the formula follows.
\end{proof}

\noindent\textbf{2.}
The presented theorem reduces the problem of classification of
finite-dimensional irreducible representations $(C^2, U)$
to the description of cycles of the dynamical system
${\mathbb{R}^{1}}\stackrel{F(\cdot)}{\to}{\mathbb{R}^{1}}$.
So, let us look how the facts about cycles of
dynamical systems can be used in the
context of the representation theory.

Sharkovsky's theorem establishes the following order in the set of natural
numbers.

\begin{theorem}
Let $F\colon I\mapsto I$ be a continuous mapping of the closed
interval $I$ into itself. If the dynamical system possesses a
cycle of period $m$, then for any $m' \triangleleft m$,
there exists a cycle of period $m'$, where $\triangleleft$
denotes the following order on the set $\mathbb{N}$ of
natural numbers\textup:
\begin{equation}
1 \triangleleft 2 \triangleleft \dots \triangleleft 2^{n}
\triangleleft \cdots \triangleleft
2^{2} \cdot 5 \triangleleft 2^{2} \cdot 3 \triangleleft \cdots
\triangleleft 2 \cdot 5 \triangleleft 2 \cdot 3
\triangleleft \dots  \triangleleft 5
\triangleleft 3.
\end{equation}
For any $m$, there exists a continuous mapping $\tilde F\colon I
\mapsto I$ such that the dynamical system has a cycle of period $m$,
and does not have cycles of periods $m'$ for  $m \triangleleft m'$.
\end{theorem}

This theorem gives the following statement about the dimensions of
irreducible representations of \eqref{xx}.

\begin{proposition}
Let $F\colon I\mapsto I$ be a continuous mapping of the interval $I$
into
itself. If there exist irreducible $m$-dimensional representations
of \eqref{xx} such that $\sigma(C^2) \subset I$, then
for any $m'\triangleleft m$,  there exist $m'$-dimensional
irreducible
representations of \eqref{xx}.

For any $m$, there exists a continuous mapping $\tilde F\colon I
\mapsto I$
such that the relation \eqref{xx} has  an $m$-dimensional irreducible
representation and does not have irreducible representations of
dimension
$m'$ for  $m \triangleleft m'$.
\end{proposition}

For continuous mappings $F(\cdot)$,  the following corollaries hold.

\begin{corollary}
If   relation \eqref{xx} with continuous $F(\cdot)$ has irreducible
representations with a dimension which is
not equal to a power of\/ $2$, then there are infinitely many
dimensions for
which \eqref{xx} has irreducible representations.
\end{corollary}

\begin{corollary}
The
existence of a three-dimensional irreducible representation of
\eqref{xx}
implies that \eqref{xx} has irreducible representations of any
dimension
$n\in \mathbb{N}$.
\end{corollary}

\begin{example} (Second-degree mapping. Finite-dimensional
representations).
Consider the following relation
\[
xx^* = (x^*x -q)^2.
\]
The corresponding dynamical system is generated by the polynomial
$P_q(\lambda)=(\lambda -q)^2$.
According to the arguments
above, all finite-dimensional representations are described in terms of
cycles of this mapping. Let us look at how the value of $q$ affects the
existence of cycles and their order (see, e.g., \cite{shmr})

For $q<-1/4$, there are no stationary points, and therefore, there are
no cycles at all. For $q=-1/4$, there exists a unique stationary point,
$\lambda=1/4$, and no other cycles. To this point, there corresponds a
circle of one-dimensional representations, $X= e^{i\phi}/2$, $\phi \in
[0,2\pi)$.

For $-1/4 < q <3/4$, there are two stationary points, $\lambda_{0,1}
= (2q +1
\pm \sqrt{4q +1})/2$, which give two one-dimensional families of
irreducible representations. There are no other cycles and no other
irreducible finite-dimensional representations.

As $q$ increases from 3/4 to $q^* \approx 1.4$, cycles of order 2,
$2^2$, \dots, $2^n$, and the corresponding families of irreducible
representations of the corresponding dimensions arise.

For $q=q^*$, there exist cycles of any order $2^k$, $k\ge 1$, and no
other cycles; any irreducible finite-dimensional representation has
dimension $2^k$ for some $k\ge0$.

As $q>q^*$ increases, other cycles arise in the order described by
the Sharkovsky theorem. Starting from some $q\approx
1.75$,  there are cycles of order 3,
and therefore, cycles of all
other orders. Therefore, for such $q$
there are irreducible representations of any finite
dimension. Notice that for some values of $q$, zero may become periodic
point (e.g., for $q=1$, zero is a periodic point of the second order);
in this case, the corresponding one-parameter family of representations
degenerates into a single irreducible representation.


Relation \eqref{rel:quadratic}
\[
xx^* = \alpha x^*x(I- x^*x)
\]
has a
similar set of finite-dimensional representations, but the
family corresponding to the fixed
point $\lambda =0$ degenerates into the unique trivial
representation, and there is no degeneration of representations
corresponding to other cycles.
The corresponding critical values of $\alpha$ are: $\alpha=3$
(two-dimensional representations arise), $\alpha=\alpha^*\approx
3.569$ (there are
representations with dimensions of any power of 2, and no others),
$\alpha\approx
3.8$ (there is a three-dimensional representation, and thus
representations with any dimensions).
\end{example}

\begin{example}(Continuous fractions. Finite-dimensional representations).
Consider operator relations which arise from  the M\"obius
mapping
\begin{gather}\label{moebius-rel}
xx^* = (a x^*x +c)(bx^*x +d)^{-1},\notag
\\
a,b,c,d \in \mathbb{R},\quad
b >0, \quad ad -bc \ne 0.
\end{gather}
According to Theorem~\ref{fin_th}, in order to describe finite-dimensional
representations of relation \eqref{moebius-rel}, one needs to find cycles of
the dynamical system generated by the mapping
\begin{equation}\label{moebius-map}
F(z) = \frac{az +c}{bz +d}.
\end{equation}
To do that, we follow \cite{83}. First,
consider fixed points of the mapping. If $(d-a)^2 +4bc =0$, then there exists a
single fixed point $\xi_1=(a-d)/2b$;
otherwise, there are two fixed points, $\xi_1$,
$\xi_2$.

If there exists a single stationary point $\xi_1$, then $F$ is conjugate to the
shift in the complex plane, $F = \phi^{-1} \circ T \circ \phi$, where $\phi(z)=
1/(z-\xi_1)$, $T(w) = w+l$, $l = 2b/(a+d)$. We see that, in this case, $\xi_1$ is
a unique periodic point. Representations exist only if $\xi_1\ge0$;
for  $a=d$, $X=0$ is a single solution of
\eqref{moebius-rel}, for $\xi_1>0$ there exists a one-parameter family of
one-dimensional representations, $X = \alpha \sqrt{\xi_1}$, $|\alpha| =1$.

If there are two stationary points, $\xi_1$, $\xi_2$, one can construct
one-dimensional representations quite similarly, if one  or
both of these points
are non-negative.

If there are no stationary points (they are conjugate complex-valued),
write
\[
\phi(z) = \frac{z-\xi_1}{z-\xi_2};
\]
then $F = \phi^{-1} \circ T \circ \phi$, where $Tw = qw$ with $q =
(a-\xi_1b)/(a-\xi_2 b) = (d - \xi_2 b)/(d-\xi_1b)$.  Now
it is easy to see that
either $q$ is the $n$-th root of the unity for some $n$,  and
all points are periodic
with period $n$, or there are no periodic points at  all.
In the periodic case,
representations correspond to orbits with all
non-negative points, and these
representations are constructed according  to Theorem~\ref{fin_th}.

Notice that in this example the rule about dimensions of representations
established by the Sharkovsky theorem does not hold: there can be only
one-dimensional and $n$-dimensional irreducible representations.
\end{example}

\subsection{Infinite-dimensional representations}\label{sec:2.1.3}

In order to describe the general case, recall that, according to
Proposition~\ref{xx-center}, the operator $U$ is a centered partial
isometry. We will see that in the irreducible representation with
non-unitary $U$, the pair $U$, $U^*$ is again irreducible; since all
irreducible partial isometries can easily be described, this enables us
to give a complete description of all irreducible representations in
the non-unitary case.

In the unitary case, two classes of representations can arise:
representations in which $U$ acts as a shift in $l_2$ (in this case
the spectrum of $C^2$ lies on a single orbit), and representations
corresponding to non-trivial ergodic measures. The latter class of
representations is too complicated to be classified up to a unitary
equivalence
for the moment; however,
non-trivial ergodic measures can arise only if the corresponding
dynamical system does not have a measurable section.

\medskip\noindent\textbf{1.}
We start with the description of centered partial isometries.

\begin{theorem}\label{thiso-x}
Any irreducible centered partial isometry is
one of the following:

\begin{itemize}
\item[\textup{(i)}]
 one-dimensional unitary operator $U=\alpha $, $|\alpha
|=1;$

\item[\textup{(ii)}]  unilateral shift operator in $l_2$,
$Ue_k=e_{k+1}$\textup;

\item[\textup{(iii)}]
 adjoint to the unilateral shift operator in $l_2$, $Ue_k = e_{k-1}$,
$k>1$, $Ue_1=0$\textup;

\item[\textup{(iv)}]  finite-dimensional operator in $\mathbb
C^n$ of the form $Ue_k=e_{k+1}$,
$k=1,\dots ,n-1$, $Ue_n=0$, for some $n=1$, $2$, \dots  .
\end{itemize}
\end{theorem}

\begin{proof}
We start with a simple fact.

\begin{proposition}
The operators $U^k(U^{*})^k$, $(U^{*})^lU^l$, $k$, $l=1$, $2$,\dots ,
are projections.
\end{proposition}

\begin{proof}
Since $UU^{*}U=U$ and the operators $U^{*}U$ and
$U^{k-1}(U^{*})^{k-1}$ commute, we have by induction that
\begin{align*}
U^k(U^*)^kU^k(U^*)^k&=
UU^{k-1}(U^*)^{k-1}U^*UU^{k-1}(U^*)^{k-1}U^*\\
&=UU^*UU^{k-1}(U^*)^{k-1}U^{k-1}(U^*)^{k-1}U^*\\
&=UU^{k-1}(U^*)^{k-1}U^*=U^{k}(U^*)^{k}.
\end{align*}
Similarly, since $U^{*}UU^{*}=U^{*}$ and $UU^{*}$ and
$(U^{*})^{k-1}U^{k-1}$ commute, we get that $(U^{*})^kU^k$ is a
projection.
\end{proof}

Denote these projections by $P_k=(U^{*})^kU^k$,
$P_{-k}=U^k(U^{*})^k$, $ k=1$, $2$, \dots ; $P_0=I$.

\begin{proposition}
For all $k\in \mathbb Z$, the following relations hold
\begin{equation}
\label{iso2-x}P_kU=UP_{k+1}.
\end{equation}
\end{proposition}

\begin{proof}
Indeed, for $k>0$, we have
\begin{align*}
P_{k}U &= (U^*)^{k}U^{k} U= (U^*)^{k}U^{k}UU^*U=
UU^*(U^*)^{k}U^{k}U
\\ 
&= U(U^*)^{k+1}U^{k+1}=UP_{k+1},
\\
P_{-k}U&=U^k(U^*)^k U = UU^{k-1}(U^*)^{k-1}U^*U
\\ 
&=
UU^*UU^{k-1}(U^*)^{k-1}= UU^{k-1}(U^*)^{k-1}=UP_{-k+1}
\end{align*}
and
$
P_{-1}U=(UU^*)U=UI=UP_0$,  $P_0U=IU=U(U^*U)=UP_1$.
\end{proof}

The case where $U$ is unitary is trivial. Suppose that the
operator $U^*$ has a non-trivial kernel (the case of nontrivial
kernel of $U$ is similar). Let $f\in\ker U^*$. For every
$k=1$, $2$, \dots, consider the vector $(U^*)^kU^kf$. The following
situations may occur:

a)  $(U^*)^kU^kf=f$ for all $k=1$,  2, \dots. Then the vector
$f_0=f$ is a joint eigenvector of a commuting family $(P_k)$.

b) for some $k>0$, the following conditions hold:
\begin{gather*}
(U^*)^lU^lf=f,\qquad l=1,\dots,k-1, \\ (U^*)^kU^kf\ne f.
\end{gather*}
Put $f_0=f-(U^*)^kU^k f\ne0$. Then $(U^*)^kU^kf_0=0$, which
implies that $U^kf_0=0$, \ $U^{k+1}f_0=0$, etc., and $f_0$ is a joint
eigenvector of the commuting family $(P_k)$.

In both the cases, the relation \eqref{iso2-x} implies that
$f_0$, $Uf_0$, $U^2f_0$, \dots, are orthogonal joint eigenspaces of
the family $(P_k)$ and can be chosen to be a basis of the space.
The rest of the proof is obvious.
\end{proof}

\noindent\textbf{2.}
To apply this theorem to the description of irreducible
solutions of \eqref{xx}, we need the following fact.

\begin{theorem}\label{thirr-x}
Let the pair $(X,X^*)$ satisfying \eqref{xx} be irreducible. If one
of the operators $X$, $X^*$ has a non-zero kernel, then the pair
$(U,U^*)$ is irreducible, i.e., any bounded operator commuting with
$U$ and $U^*$ is a multiple of the identity.
\end{theorem}

\begin{proof}
First, consider the case $\ker U \ne 0$. Take a vector $e_0 \in \ker U$,
and consider vectors $e_k=(U^*)^k$, $k =1$, $2$, \dots. Since $\ker U =
\ker C$, each $e_k$ is an eigenvector of $C^2$ with  the
eigenvalue $F^k(0)$.

We show that these vectors form a basis in the space. Indeed, the linear span
$\tilde H$ of
these vectors is invariant with respect to $U^*$ and $C^2$.

Take a
vector $e_k\ne0$ with some $k>0$, and assume that $\tilde e_{k-1} = Ue_k =
UU^*e_{k-1}$
is not in $\tilde H$. Since $U^*UU^* =U^*$, we get $U^*\tilde
e_{k-1} =e_k$; therefore, $(\tilde e_{k-1} - e_{k-1})
\in \ker U^* = \ker F(C^2)$, which is the image of the projection
$E_{C^2}(F^{-1}(0))$.
On the other hand, $(\tilde e_{k-1} -
e_{k-1})$ belongs to $E_{C^2}(\Delta)H$, where
$\Delta =F^{-1}(F^k(0))$. If the latter vector is non-zero, this
would imply that both $e_{k-1}$ and $\tilde e_{k-1}$
belong to the kernel of $U^*$ and that $e_k =0 $. Therefore, $\tilde
e_{k-1}= e_{k-1}$, and $\tilde H =H$. The operator $U$ is adjoint to the
unilateral shift in $l_2$.

Now consider the case $\ker U^* \ne 0$.
For each $k\ge0$, introduce the subspace $H_k = E_{C^2}(F^{-k}(0))H$.
The equalities
\begin{align}\label{rel:euue}
UE_{C^2}(\Delta)& =E_{C^2}(F^{-1}(\Delta)) U, \notag
\\
E_{C^2}(\Delta) U^*& = U^* E_{C^2}(F^{-1}(\Delta))
\end{align}
 for all measurable $\Delta$ imply that $U H_k = H_{k+1}$, $U^*H_{k+1} =
H_k$, $k\ge1$, and the span of these subspaces is an invariant subspace; due
to the irreducibility,  it is the whole of $H$.

Note that $\ker U^* = \ker F(C^2)=H_1$.
First we show that $H_1$ is an eigenspace of $C^2$, i.e., there
exists a single point $\lambda_0$ of the spectrum of $C^2$
such that $F(\lambda_0)=0$. Indeed, take any measurable subset $\delta
\subset F^{-1}(0)$. Consider the subspaces $H^\delta_k =
E_{C^2}(F^{-k}(\delta))H\subset H_{k+1}$,
$k\ge0$. Equalities \eqref{rel:euue}
imply that the span of the subspaces $H_k^\delta$ is an invariant
subspace; the irreducibility then implies that there exists a single point
$\lambda_0$, $F(\lambda_0) =0$, such that
$E_{C^2}(F^{-1}(0) \setminus \lambda_0) =0$.

Thus, $H_1$ is an eigenspace of $C^2$. Now we show that $H_2$ is also an
eigenspace. Since $UU^*$ commutes with $F(C^2)$, $UU^*$ maps $H_2$
into itself, and is a projection on it. Actually, since $\ker U^* = H_1$,
we conclude that $UU^*$ is the identity on $H_2$. Now we aply similar
arguments as above. Take any measurable $\delta \subset
F^{-1}(\lambda_0)$, and consider the subspaces $H_k^\delta =
E_{C^2}(F^{-k}(\delta)) \subset H_{k+2}$, $k\ge 0$. Relations
\eqref{rel:euue}  and $UU^*\restriction H_2=I$ imply that the span of
$H_k^\delta$, $k \ge0$, and $H_1$ is an invariant subspace; therefore,
by the irreducibility, $H_2$ is also an eigenspace of $C^2$.

Continuing this process, we conclude
that all $H_k$, $k\ge1$, are eigenspaces
of $C^2$. Also, in the irreducible case, all these  eigenspaces are
one-dimensional; therefore, $U$ is a unilateral shift in $l_2$.
\end{proof}

\noindent\textbf{3.}
Now we apply the results on centered partial isometries to the description
of  operators satisfying \eqref{xx} (note
that the same result can be obtained
using \cite{vai}).
 Since we are interested in infinite-dimensional
representations, only isometries, or co-isometries may arise in
the infinite-dimensional non-unitary case.


\begin{theorem}\label{thdeg-x}
Any infinite-dimensional irreducible
representation of relations \eqref{xx},
for which $\ker X\cup\ker X^*\ne\{0\}$, falls into one of the following
classes\textup:
\begin{itemize}
\item[\textup{(i)}] infinite-dimensional in $l_2$\textup:
\begin{gather*}
X e_1 =0, \quad Xe_j=\sqrt{\lambda_j}e_{j-1},\qquad j=2,3,\dots,
\\
 \lambda_j>0, \quad
\lambda _j = F^{j-1}(0)
\end{gather*}
\textup(Fock representation\textup),

\item[\textup{(ii)}] infinite dimensional in $l_2$\textup:
\begin{gather*}
Xe_j=\sqrt{\lambda_j}e_{j+1},\qquad j=1,2,\dots,
\\
\lambda_j>0, \quad \lambda_j \in F^{-j}(0),\quad 
F(\lambda_{j+1}) = \lambda_j
\end{gather*}
\textup(anti-Fock representations\textup).
\end{itemize}
\end{theorem}

\begin{proof}
Indeed, the phase $U$ of the operator $X$
is a partial isometry (non-unitary), thus by Theorem~\ref{thirr-x},
the representation space of an irreducible $X$ is the same as
for an irreducible $U$. Use Theorem~\ref{thiso-x} to represent the
operator $U$; the rest of the proof follows immediately from
\eqref{cu}.
\end{proof}

\begin{example}(Mapping of degree two. Fock and anti-Fock representations).
For relation \eqref{rel:quadratic}, zero is a stationary point; therefore,
this relation admits no Fock or anti-Fock representations.

However, this is not true for \eqref{rel:parabola}; below, we look at such
representations for the relation
\begin{equation}\label{rel:parabola-x}
xx^* = (x^*x -q I).
\end{equation}
As follows from the previous theorem, we need to consider half-orbits
lying in the positive region, and coming from or going to zero.

1. Fock representations. Image of zero under the action of powers of $F$
is always positive except for the case $q=0$. Therefore, for any $q\ne
0$, one can construct the Fock representation of \eqref{rel:parabola-x}.
However, for $q< -1/4$, the sequence $F^n(0)$ is unbounded, and therefore,
the corresponding representation is unbounded. For all $q> -1/4$, for
which zero is not a periodic point, there exists a unique bounded
Fock representation of \eqref{rel:parabola-x}; the spectrum of $C^2$ lies
between zero and the first stationary point for $q<0$, and on
the interval $[0,q^2)$ for $q \in (0,2]$. Notice that for $q=1$, zero is a
periodic point of the second order, and the corresponding representation
is two-dimensional. Similarly, for those values of\/
$q$ for which $q$ is a
periodic point of some order $n$, the corresponding Fock representation is
$n$-dimensional.

For $q>2$, the Fock representation is unbounded.

2. Anti-Fock representations. Now we consider positive half-orbits
going
into zero. It is obvious that this is possible only for $q>0$. For $0<q<1$,
there exists a unique sequence of positive pre-images
of zero, $F^{-k}(0) \to
\lambda_2$, where $\lambda _2$ is the second
stationary point of the mapping.

For $q=1$, zero is a periodic point of period 2, and zero has two positive
pre-images, 0 and 2. However, the representation that correspond to 0 is
two-dimensional, and we again have a single infinite-dimensional anti-Fock
representation. Indeed, one can easily see that there are at least
countably many inverse orbits.

For $q>1$, zero has two positive pre-images,
$0<t_1<q<t_2<\lambda_2$, which generate at least countably many
sequences of positive pre-images.

\begin{proposition}
For $1<q\le q^*$ there are countably many inequivalent irreducible
anti-Fock representations. For $q>q^*$, there is a continuum of such
representations.
\end{proposition}

\begin{proof}
First mark some points on the line. Denote by $\lambda_1$, $\lambda_2$,
$\lambda_1< \lambda_2$, the stationary points of $F(\cdot)$;
$\lambda_1'\ne
\lambda_1$ is the second pre-image of $\lambda_1$, $F(\lambda_1')=
\lambda_1$, and $\lambda_1''<\lambda_1$ is a different from $\lambda_1$ and
$\lambda_1'$ pre-image of $\lambda_1$ with respect to $F^2(\cdot)$,
$F^2(\lambda_1'')= \lambda_1$. Also introduce $\lambda_{\textstyle*}
= q^2 = F^2(0)$.


Let $q\le q^*$. Then $F(\cdot)$ has cycles only of orders $2^k$, $k\ge
1$.
$F$ maps the  interval $[0,\lambda_{\textstyle*}]$
into itself, and one can easily see
 that
the number of inverse orbits that leave this interval does not influence 
the 
total cardinality of the inverse orbits, provided it is infinite. Thus, 
we need to investigate the number of orbits that go to zero and lie 
completely inside $[0,\lambda_{\textstyle *}]$.

Consider two intervals, $I_1^1 = [\lambda_1'',\lambda_1]$ and $I_2^1 =
[\lambda_1, \lambda_1']$. They are invariant with respect to $F^2(\cdot)$.
Indeed, it is sufficient to show that $\lambda_1' > \lambda_{\textstyle *}$; if not,
the dynamical system $F^2$ on the interval $I_2^1$
would have the same
dynamics  as
the quadratic mapping in the case of $q\ge2$ \cite{sh_kol_etal},
and $F^2(\cdot)$ would have
cycles of any period, which contradicts the condition $q\le q^*$.
Notice also that in this case, $\lambda_1'' <0$. Therefore, there are no
orbits of $F^2(\cdot)$ which walk both inside $I_1^1$ and $I_2^1$.


Now we investigate the set of orbits of the mapping $F^2(\cdot)$ on each
of the intervals, $I_1^1$
and $I_2^1$. On each of these intervals, $F^2(\cdot)$
has similar dynamical properties as a quadratic mapping; in particular,
for $F^2(\cdot)$ there are cycles only of orders $2^k$, $k \ge 1$.
Again, there are four intervals, $I_k^2$, $k=1$, \dots, 4, which are
invariant with respect to $F^4(\cdot)$, and which determine the
cardinality of orbits. Continuing this process, we get either a finite number
of intervals where $F^n(\cdot)$  has only one orbit going to
zero (for $q<q^*$),
or a Cantor set on which $F(\cdot)$ is one-to-one (for $q  = q^*$);
in both cases, we have a countable number of orbits going to zero.


To prove that for $q>q^*$ there is a continuum of anti-Fock
representations, consider first the case of $q=q_0\approx 1.54$, such that
$\lambda_{\textstyle *} =\lambda_1'$, i.e., $F^2(0) = \lambda_1$. In this case, the
intervals $I_1^1$ and $I_2^1$ are invariant with respect to
$F^2(\cdot)$, and on each of these intervals, $F^2(\cdot)$ has the
same dynamics as the quadratic mapping with $q=2$. One can easily see
that in this case the set of pre-images of zero under $F^2(\cdot)$
contains an infinite binary tree, and therefore, an uncountable number of paths
in it.

For an arbitrary $q>q^*$, one can show using \cite{sh_kol_etal} that some
iteration of $F(\cdot)$ possesses the same property, and therefore, there
is an uncountable number of orbits passing through zero.
\end{proof}
\end{example}

\begin{example}(Two-parameter unit quantum disk). For
relations related to continuous fractions, we consider only a very
special case of the mapping
\begin{equation} \label{f:qdisk}
F(\lambda) = \frac{(q+\mu) \lambda + 1- q - \mu}{\mu\lambda +1 -\mu},
\end{equation}
with $0 \le \mu \le 1$, $0 \le q \le 1$, $(\mu,q) \ne (0,1)$. This mapping
is related to the introduced in  \cite{klles2} two-parameter unit quantum disk
algebra, which is generated by the relation
\[
qzz^* - z^*z = q-1 + \mu(1-zz^*)(1-z^*z).
\]
Putting $x=z^*$, we come to relation of the form \eqref{xx} with the
such
introduced $F(\cdot)$. If $\mu=0$, we get the $q$-CCR relation considered
above;
thus we assume that
$\mu \ne 0$.


The mapping $F(\cdot)$ is one-to-one and possesses
two stationary points, $t_1 =
1-({1-q})/\mu$ and $t_2=1$, that correspond to two families of
one-dimensional representations.
Consider the following cases.

1. Let $q=1$. In  this case, $t_1 = t_2$, and besides the
one-dimen\-sional
family, there exists a unique infinite-dimensional bounded
representation that
corresponds to the sequence of the pre-images of zero, $\lambda_k =
F^{-k}(0)=  \mu k /(1+\mu k)$, $k\ge1$.
The operator $z$ is
\begin{align*}
ze_k &= \sqrt{\lambda_{k-1}}\,e_{k-1},
\\
z^*e_k& = \sqrt{\lambda_k}\, e_{k+1}, \qquad k \ge 1.
\end{align*}

2. Let $1-\mu < q <1$. Now there are two one-dimensional families, and the
representation corresponding to the sequence of pre-images of zero,
$\lambda_k = F^{-k}(0)=(1-q^{-k}+\mu q^{-(k+1)/2}[k]_{\sqrt q})/(1+\mu 
q^{-(k+1)/2}[-k]_{\sqrt q})$,
$k \ge1$ (we use the notation $[n]_q = (q^n - q^{-n})/(q-q^{-1})$). The
representation acts by the same formula with so introduced $\lambda_k$.
But besides these representations, there exists another family of bounded
representations such that the kernels of $z$ and $z^*$   are
zero (see below).

3. For $q=1-\mu$, zero is a fixed point,  which corresponds to the trivial
representation. In this case, there exists a family of bounded
representations with zero kernels of $z$ and $z^*$ (see below).

4. For $0\le q < 1-\mu$, the value $t_1$ is negative, and there is again
only one family of one-dimensional representations corresponding to the
fixed point $t_2=1$, and a single bounded representation corresponding
to the
sequence of images of zero, $\lambda_k = F^k(0) =(1-q^k
- \mu  q^{(k-1)/2} [k]_{\sqrt q})/(1-\mu  q^{(k-1)/2} [k]_{\sqrt q})$,
$k \ge0$,
\begin{align*}
ze_k & =\sqrt{\lambda_k}\, e_{k+1},
\\
z^*e_k & =\sqrt{\lambda_{k-1}}\, e_{k-1}, \qquad k\ge1.
\end{align*}
\end{example}

\noindent\textbf{4.}
Now, it remains to consider the case of the unitary operator $U$. We
distinguish between two cases: representations related to a single
orbit, and representations related to a non-trivial ergodic
quasi-invariant measure.

If there exists a measurable section of the dynamical system, then any
ergodic measure is concentrated on a single orbit; in this case we will
provide a complete description of the representations.

\begin{theorem}
Let the operator $X$ be invertible, and the dynamical system possess
a measurable section. Any
infinite-dimensional irreducible representation of \eqref{xx} has the
form
\begin{equation}\label{rel:unitary}
Xe_k = \sqrt{\lambda_k}\, e_{k-1}, \qquad k \in \mathbb{Z},
\end{equation}
where $\lambda_k$, $k\in \mathbb{Z}$, is any sequence of positive
numbers such that $F(\lambda_k) = \lambda_{k+1}$, $k \in \mathbb{Z}$.
Two such representations are unitarily equivalent if and only if the
corresponding sequences coincide.
\end{theorem}

\begin{proof}
Let first the mapping $F(\cdot)$ be one-to-one on the spectrum of $C^2$.
In this case, any ergodic quasi-invariant measure is concentrated on a
single orbit of the dynamical system, and is equivalent to an atomic
measure concentrated at points of the orbit. Choose any point $\lambda$
of the orbit, and take a unit eigenvector $e_0$ corresponding to
$\lambda$. Write $e_k= U^ke_0$, $k\in \mathbb{Z}$. We claim that the
space spanned by these vectors is invariant under $X$ and $X^*$.
For $k<0$, we have
\begin{align*}
C^2 e_k &= C^2 U^{k} e_0 = C^2 {U^*}^{|k|} e_0
\\
& = {U^*}^{|k|} F^{|k|}(C^2)\, e_0 = {U^*}^{|k|} F^{|k|}(\lambda)\, e_0
= F^{|k|}(\lambda)\, e_k.
\end{align*}
Also, since $F(\cdot)$ is one-to-one, we
have in this case that $C^2 U = U F^{-1}(C^2)$, which implies for each
$k>0$:
\begin{align*}
C^2e_{k}& = C^2 {U} e_{k-1} = \dots
\\
&=  C^2 U^k e_0 = U^k
F^{-k}(C^2)\, e_0 = U^k F^{-k}(\lambda)\, e_0
\\
&= F^{-k}(\lambda)\, e_k.
\end{align*}
Therefore, all $e_k$ are eigenvalues of $C^2$, and hence, of $C$. In
the chosen basis, the operator $X$ acts as stated in the theorem.

In the general case, consider a commuting family of self-adjoint
operators, $C_k = U^k C^2 U^{-k}$. The relations are $C_k U^* = U^*
C_{k+1}$. Now we have a one-to-one action of $\mathbb{Z}$ on the
infinite-dimensional space $\mathbb{R}^\mathbb{Z}$. Using the same
arguments as above, one concludes that the joint spectrum of the
commutative
family is concentrated on a single orbit, the representation space is
generated by $\delta$-functions concentrated at points of the orbit,
$U$ acts as a shift, and the formula follows.
\end{proof}

\begin{remark}
Notice that we do not assume here that the dynamical system is one-to-one.
Looking closely at the spectrum of\/ $C^2$, one sees that it can 
just be a chain of points, or a chain ``glued'' to a cycle or stationary
point. In the case of the mapping on the spectrum being one-to-one, only
chains may arise.
\end{remark}


\begin{example}
(Mapping of degree two. Continued: $q<q^*$).
Since for $q<q^*$, relation \eqref{rel:parabola} possesses only a finite
number of cycles, the set of periodic points is closed; therefore, the
corresponding dynamical system is ``simple'', and therefore, all
infinite-dimensional irreducible
representations with unitary $U$ are described by sequences of positive
numbers $\lambda_k$, $k \in \mathbb{Z}$, such that $F(\lambda_k) =
\lambda_{k+1}$ for all $k$ according to \eqref{rel:unitary}.
\end{example}

\begin{example} (Two-parameter unit quantum disk.
Non-degenerate case).
For $1-\mu \le q <1$, the dynamical system \eqref{f:qdisk} possesses a
family of bounded positive orbits. Take $\lambda_0 \in (t_1, 1)$ as an
initial point; then $H= l_2(\mathbb{Z})$, and
\begin{align*}
ze_k & = \sqrt{\lambda_{k-1}}\, 
e_{k-1},
\\
z^*e_k & = \sqrt {\lambda_k}\,
e_{k}, \qquad k \in \mathbb{Z}.
\end{align*}
where
\[
\lambda_k =   \frac{\lambda_0 (q^{k} + \mu q^{(k-1)/2} [k]_{\sqrt q}) + 1-
(q^{k} + \mu q^{(k-1)/2} [k]_{\sqrt q})}{\lambda_0 \mu q^{(k-1)/2}
[k]_{\sqrt q}
+ 1 - \mu q^{(k-1)/2} [k]_{\sqrt q}}.
\]
Two such representations corresponding to $\lambda_0$ and $\lambda_0'$
are unitarily equivalent if and only if $\lambda_0$ and $\lambda_0'$
are on
the same orbit. Therefore, for the measurable section one can take any
interval of the form $\bigl[\lambda_0,
\frac{(q+\mu)\lambda_0 + 1 -q - \mu}{\mu
\lambda_0 +1 -\mu}\bigr)$.

Also, there exists a family of unbounded positive orbits lying to the right of 1.
The set of such orbits also possesses a measurable section, and one can
use the formula above to construct the corresponding class of irreducible
representations. However, these representations are unbounded.
\end{example}


If the dynamical system does not have a measurable section, the
representations listed in the theorem above do not form, in general,
the complete list of irreducible representations.

\begin{example} (Mapping of degree two. $q\ge q^*$).
For any $q \ge q^*$, there exists an invariant set $K$ homeomorphic to
a Cantor set where the mapping is one-to-one and which carries a
non-trivial ergodic quasi-invariant probability measure $\mu$.
Consider the space $L_2(K, d\mu)$, and define $X$ by
\begin{align*}
(Xf)(\lambda)& = \lambda \sqrt{d\mu(F(\lambda))/d\mu(\lambda)}\,
f(F(\lambda)),
\\
(X^*f)(\lambda) & = F^{-1}(\lambda )\sqrt{d
\mu(F^{-1}(\lambda))/d\mu(\lambda)}\, f(F^{-1}(\lambda)),
\end{align*}
where $ F(\lambda) =
(\lambda -q)^2$. Then
the operators satisfy the relation $XX^* = (X^*X - qI)^2$, the
representation is irreducible, but any orbit has
spectral measure of $X^*X$ equal to zero.
\end{example}

\begin{remark}
Notice that the description of nontrivial
ergodic measures can be a very complicated task; also, there can be lots
of unitarily inequivalent irreducible representations corresponding to
the same non-trivial ergodic measure.
\end{remark}

%\begin{example}(Continued fractions. Infinite-dimensional
%representations)
%\end{example}

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